Q.(a) An economy produces only coal and steel. These two commodities serve as intermediate inputs in each other's production. 0.4 tonne of steel and 0.7 tonne of coal are needed to produce a tonne of steel. Similarly 0.1 tonne of steel and 0.6 tonne of coal are required to produce a tonne of coal. No capital inputs are needed. Do you think that the system is viable? 2 and 5 labour days are required to produce a tonnes of coal and steel respectively. If economy needs 100 tonnes of coal and 50 tonnes of steel, calculate the gross output of the two commodities and the total labour days required.

Use the property $|A^{-1}|=\dfrac{1}{|A|}$, giving $\dfrac{1}{5}$.

From $A\,A^{-1}=I$, taking determinants and using $|AB|=|A||B|$:

$|A|\cdot|A^{-1}|=|I|=1.$

Therefore

$|A^{-1}|=\frac{1}{|A|}.$

Since $|A|=5\neq 0$, $A$ is invertible and

$|A^{-1}|=\frac{1}{5}.$

…

Compute $|A|=\dfrac{7}{30}$ and $\text{adj}(A)$, then $A^{-1}=\dfrac{1}{|A|}\text{adj}(A)=\dfrac{30}{7}\begin{pmatrix}1/2 & 5/12\\ 2/5 & 4/5\end{pmatrix}$.

Let $A=\begin{pmatrix} \dfrac{4}{5} & \dfrac{-5}{12} \\[4pt] \dfrac{-2}{5} & \dfrac{1}{2} \end{pmatrix}$.

Step 1 — Determinant.

$|A|=\frac{4}{5}\cdot\frac{1}{2}-\left(\frac{-5}{12}\right)\left(\frac{-2}{5}\right)=\frac{4}{10}-\frac{10}{60}=\frac{2}{5}-\frac{1}{6}.$

$=\frac{12}{30}-\frac{5}{30}=\frac{7}{30}.$

Step 2 — Adjoint. For a $2\times 2$ matrix $\begin{pmatrix}a&b\\ c&d\end{pmatrix}$, $\text{adj}=\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}$:

$\text{adj}(A)=\begin{pmatrix} \dfrac{1}{2} & \dfrac{5}{12} \\[4pt] \dfrac{2}{5} & \dfrac{4}{5} \end{pmatrix}.$

Step 3 — Inverse. …

Input-output analysis, a matrix-based technique for modelling the interdependence of industries in an economy, was invented by Prof. Wassily W. Leontief.

In the Tamil Nadu HSC Class-11 Business Mathematics syllabus, the Matrices and Determinants chapter introduces input-output analysis as an application of matrices. In this model the inter-industry demand of an economy is written as a matrix equation $X = AX + D$, whose solution $X = (I - A)^{-1}D$ needs the matrix inverse.

…

The Hawkins-Simon conditions for the viability of an input-output model are two in number, so the answer is option (c) 2.

In Leontief's input-output analysis the technology matrix $B$ (or the matrix $I - A$, where $A$ is the coefficient matrix) must satisfy conditions that guarantee the system produces non-negative output levels able to meet the final demand.

The Hawkins-Simon conditions state that:

1. The main diagonal elements of $(I - A)$ must be positive, and …